Partial Fractions Decomposition-Unsure Which Method To Use When

Question

So I was working on this problem and could not use the cover up method to solve it. I was getting the wrong answer.

Find B. $$\frac{1}{s^2(s^2+4)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+4}$$ $$s=0: \frac{1}{4}=B$$ Finding C, D, and A. $$s=1:\frac{1}{5}=A+\frac{1}{4}+\frac{C+D}{5}$$ $$s=-1:\frac{1}{5}=-A+\frac{1}{4}+\frac{-C+D}{5}$$ $$s=2:\frac{1}{32}=\frac{A}{2}+\frac{\frac{1}{4}}{4}+\frac{2C+D}{16}$$ Got rid of denominators and got a system of equations. $$\begin{cases}20A+4C+4D=-1\\-20A-4C+4D=-1\\16A+4C+2D=-1\end{cases}$$ Then I used row reduction and got the following values for A, C, and D. $$A=1/8$$ $$C=-5/8$$ $$D=-1/4$$ But I was told you can only use the coverup method when the denominator is a product of linear factors. So I used the coefficient comparison method to get the correct answer.

$$A=0$$ $$B=1/4$$ $$C=0$$ $$D=-1/4$$ HOWEVER, I am not sure this is true because I used the cover up method for the following problem and got the correct answer. $$\frac{s}{(s^2+4)(s+2)}$$

So my question is what method should I use when and is there a one size fits all method to use to avoid these issues?

Answer 1

I think that there are arithmetic errors which are causing you problems.

In your first approach, you are substituting values in for $s$ in order to obtain a system of equations which the coefficients must satisfy. However, your first substitution doesn't make sense in terms of the expression you have written, as the fractions on both sides of the equation are undefined when $s = 0$. As such, it is not clear to me how you obtained the value for $B$. That being said, you could clear the denominators, then substitute $0$ for $s$, and should obtain that same result.

However, in the next step, there seems to be an error. When $s=2$, I get $$ \frac{1}{32} = \frac{A}{2} + \frac{1}{16} + \frac{2C + D}{8}. $$ Note the $8$ in the denominator of the final term, rather than a $16$.

While I have not carried out the entire computation using your start, I suspect that will fix the problem.

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If I were given the task of finding a partial fraction decomposition of the given rational function, I would proceed as follows:

\begin{align*} &\frac{1}{s^2(s^2 + 4)} = \frac{A}{s^2} + \frac{B}{s} + \frac{Cs + D}{s^2 + 4} \\ &\qquad\iff 1 = A(s^2+4) + Bs(s^2+4) + (Cs+D)s^2 && \text{(clear the denominators)} \\ &\qquad\iff 1 = (B+C)s^3 + (A+D)s^2 + 4Bs + 4A. \end{align*}

Two polynomials are equal if and only if they have the same coefficients. Writing out the coefficients on the left explicitly gives $$ \color{blue}{0} s^3 + \color{red}{0} s^2 + \color{green}{0} s + \color{violet}{1} = \color{blue}{(B+C)}s^3+\color{red}{(A+D)} s^2 + \color{green}{4B}s + \color{violet}{4A},$$ from which is follows that $$\begin{cases} \color{blue}{0} = \color{blue}{B+C}, \\ \color{red}{0} = \color{red}{A+D}, \\ \color{green}{0} = \color{green}{4B}, \\ \color{violet}{1} = \color{violet}{4A}. \\ \end{cases}$$ The third and fourth equations give $A = 1/4$ and $B = 0$. Then $$ 0 = B+C = 0 + C = C \implies C = 0, $$ and $$ 0 = A+D = \frac{1}{4} + D \implies D = -\frac{1}{4}.$$ It is therefore possible to conclude that $$\boxed{ \frac{1}{s^2(s^2+4)} = \frac{1/4}{s^2} + \frac{0}{s} + \frac{-1/4}{s^2+4} = \frac{1}{4s^2} - \frac{1}{4(s^2+4)}. }$$ Notice that if you expand out the expression on the right, you end up with $$\frac{1}{4s^2} - \frac{1}{4(s^2+4)} = \frac{(s^2+4) - s^2}{4s^2(s^2+4)} = \frac{4}{4s^2(s^2+4)} = \frac{1}{s^2(s^2 + 4)},$$ which verifies that the decomposition is correct.

Answer 2

By equalizing the numerators between left-side and right-side after the right-side was turned into one rational expression with the same denominator as the rational expression at the left-side, then: $$1=As(s^2+4)+B(s^2+4)+s(Cs+D)$$ $$0\cdot s^3+0\cdot s^2+0\cdot s^1+1\cdot s^0=(A+C)s^3+(B+D)s^2+4A\cdot s^1+4B\cdot s^0$$ $$A+C=0,B+D=0,4A=0,4B=1$$

Answer 3

Substituting $t=s^2$ it suffices to decompose $${1\over t(t+4)}={A\over t}+{B\over t+4}\quad (*)$$ In order to get $B$ multiply both sides by $t+4$ to get $${1\over t}={A(t+4)\over t}+B$$ Plugging in $t=-4$ gives $ B=-{1\over 4}.$ Similarly multiplying $(*)$ by $t$ and plugging in $t=0$ gives $A={1\over 4}.$ Summarizing $${1\over s^2(s^2+4)}={1\over 4s^2}-{1\over 4(s^2+4)}$$ Remark Consider a different expression $${1\over s^2(s+4)} ={A\over s}+{B\over s^2}+{C\over s+4}$$ First we decompose $${1\over s(s+4)}={1\over 4s}-{1\over 4(s+4)}$$ Then we multiply the result by ${1\over s}$ and decompose once more.